Chapter 4 – Principle of Mathematical Questions and Answers: NCERT Solutions for Class 11 Maths

( Exercise 4.1)

1. Prove that following by using the principle of mathematical induction for all n∈Nn∈N:
1+3+32+…..+3n-1=(3n-1)21+3+32+…..+3n-1=(3n-1)2

Ans: Let us denote the given equality by P(n),P(n), i.e.,
P(n):1+3+32+…..+3n-1=(3n-1)2P(n):1+3+32+…..+3n-1=(3n-1)2
For n=1n=1,
L.H.S.=31−1=1L.H.S.=31−1=1
R.H.S.=(31-1)2=3-12=22=1R.H.S.=(31-1)2=3-12=22=1
Therefore, P(n)P(n) is true for n=1n=1.
Let us assume that P(k)P(k) is true for some positive integer kk, i.e.,
1+3+32+…..+3k-1=(3k-1)2…..(i)1+3+32+…..+3k-1=(3k-1)2…..(i)
Now, we have to prove that P(k+1)P(k+1) is also true.
Consider
1+3+32+…..+3k-1+3(k+1)-11+3+32+…..+3k-1+3(k+1)-1
=(1+3+32+….+3k-1)+3k=(1+3+32+….+3k-1)+3k
=(3k-1)2+3k=(3k-1)2+3k Using(i)Using(i)
=(3k-1)+2.3k2=(3k-1)+2.3k2
=(1+2)3k−12=(1+2)3k−12
=3.3k−12=3.3k−12
=3k+1−12=3k+1−12
Therefore, P(k+1)P(k+1) holds whenever P(k)P(k) holds.
Hence, the given equality is true for all natural numbers i.e., NN by the principle of mathematical induction.

2. Prove the following by using the principle of mathematical induction for all n∈N:n∈N:
13+23+33+…..+n3=(n(n+1)2)213+23+33+…..+n3=(n(n+1)2)2

Ans: Let us denote the given equality by P(n),P(n), i.e.,
P(1):13+23+33+…..+n3=(n(n+1)2)2P(1):13+23+33+…..+n3=(n(n+1)2)2

For n=1n=1,
L.H.S.=13=1L.H.S.=13=1
R.H.S.=(1(1+1)2)2=(1)2=1R.H.S.=(1(1+1)2)2=(1)2=1
Therefore, P(n)P(n) is true for n=1n=1.
Let us assume that P(k)P(k) is true for some positive integer kk, i.e.,
13+23+33+…..+k3=(k(k+1)2)2…..(i)13+23+33+…..+k3=(k(k+1)2)2…..(i)

Now, we have to prove that P(k+1)P(k+1) is also true.
Consider
13+23+33+…..+k3+(k+1)313+23+33+…..+k3+(k+1)3

=(13+23+33+…..+k3)+(k+1)3=(13+23+33+…..+k3)+(k+1)3
=(k(k+1)2)2+(k+1)3=(k(k+1)2)2+(k+1)3 Using(i)Using(i)
=k2(k+1)24+(k+1)3=k2(k+1)24+(k+1)3
=k2(k+1)2+4(k+1)34=k2(k+1)2+4(k+1)34
=(k+1)2{k2+4(k+1)}4=(k+1)2{k2+4(k+1)}4
=(k+1)2{k2+4k+4}4=(k+1)2{k2+4k+4}4
=(k+1)2(k+2)24=(k+1)2(k+2)24
=(k+1)2(k+1+1)24=(k+1)2(k+1+1)24
=((k+1)(k+1+1)2)2=((k+1)(k+1+1)2)2
Therefore, P(k+1)P(k+1) holds whenever P(k)P(k) holds.
Hence, the given equality is true for all natural numbers i.e., NN by the principle of mathematical induction.

3. Prove the following by using the principle of mathematical induction for all n∈Nn∈N :
1+1(1+2)+1(1+2+3)+…..+1(1+2+3+…n)=2n(n+1)1+1(1+2)+1(1+2+3)+…..+1(1+2+3+…n)=2n(n+1)

Ans: Let us denote the given equality by P(n),P(n), i.e.,
P(n):1+11+2+11+2+3+…..+11+2+3+…n=2nn+1P(n):1+11+2+11+2+3+…..+11+2+3+…n=2nn+1
For n=1n=1,
L.H.S.=11=1L.H.S.=11=1
R.H.S.=2⋅11+1=1R.H.S.=2⋅11+1=1
Therefore, P(n)P(n) is true for n=1n=1.
Let us assume that P(k)P(k) is true for some positive integer kk, i.e.,
1+1(1+2)+1(1+2+3)+…..+1(1+2+3+…+k)=2k(k+1)……(i)1+1(1+2)+1(1+2+3)+…..+1(1+2+3+…+k)=2k(k+1)……(i)
Now, we have to prove that P(k+1)P(k+1) is also true.
Consider
1+11+2+11+2+3+….+11+2+3+…+k+11+2+3+…+k+(k+1)1+11+2+11+2+3+….+11+2+3+…+k+11+2+3+…+k+(k+1)
=(1+11+2+11+2+3+…+11+2+3+…+k)+11+2+3+…+k+(k+1)=(1+11+2+11+2+3+…+11+2+3+…+k)+11+2+3+…+k+(k+1)
=2kk+1+11+2+3+…+k+(k+1)=2kk+1+11+2+3+…+k+(k+1) Using(i)Using(i)
=2kk+1+1((k+1)(k+1+1)2)=2kk+1+1((k+1)(k+1+1)2) [1+2+3…+n=n(n+1)2][1+2+3…+n=n(n+1)2]
=2kk+1+2(k+1)(k+2)=2kk+1+2(k+1)(k+2)
=2(k+1)(k+1k+2)=2(k+1)(k+1k+2)
=2(k+1)(k2+2k+1k+2)=2(k+1)(k2+2k+1k+2)
=2(k+1)[(k+1)2k+2]=2(k+1)[(k+1)2k+2]
=2(k+1)(k+2)=2(k+1)(k+2)
Therefore, P(k+1)P(k+1) holds whenever P(k)P(k) holds.
Hence, the given equality is true for all natural numbers i.e., NN by the principle of mathematical induction.

4. Prove the following by using the principle of mathematical induction for all n∈Nn∈N:
1.2.3+2.3.4+…+n(n+1)(n+2)=n(n+1)(n+2)(n+3)41.2.3+2.3.4+…+n(n+1)(n+2)=n(n+1)(n+2)(n+3)4

Ans: Let us denote the given equality by P(n),P(n), i.e.,
P(n):1.2.3+2.3.4+…+n(n+1)(n+2)=n(n+1)(n+2)(n+3)4P(n):1.2.3+2.3.4+…+n(n+1)(n+2)=n(n+1)(n+2)(n+3)4
For n=1n=1,
L.H.S.=1.2.3=6L.H.S.=1.2.3=6
R.H.S.=1(1+1)(1+2)(1+3)4=1.2.3.44=6R.H.S.=1(1+1)(1+2)(1+3)4=1.2.3.44=6
Therefore, P(n)P(n) is true for n=1n=1.
Let us assume that P(k)P(k) is true for some positive integer kk, i.e.,
1.2.3+2.3.4+…+k(k+1)(k+2)=k(k+1)(k+2)(k+3)4…(i)1.2.3+2.3.4+…+k(k+1)(k+2)=k(k+1)(k+2)(k+3)4…(i)
Now, we have to prove that P(k+1)P(k+1) is also true.
Consider
1.2.3+2.3.4+…+k(k+1)(k+2)+(k+1)(k+2)(k+3)1.2.3+2.3.4+…+k(k+1)(k+2)+(k+1)(k+2)(k+3)
={1.2.3+2.3.4+…+k(k+1)(k+2)}+(k+1)(k+2)+(k+3)={1.2.3+2.3.4+…+k(k+1)(k+2)}+(k+1)(k+2)+(k+3)
=k(k+1)(k+2)(k+3)4+(k+1)(k+2)(k+3)=k(k+1)(k+2)(k+3)4+(k+1)(k+2)(k+3) Using(i)Using(i)
=(k+1)(k+2)(k+3)(k4+1)=(k+1)(k+2)(k+3)(k4+1)
=(k+1)(k+2)(k+3)(k+4)4=(k+1)(k+2)(k+3)(k+4)4
=(k+1)(k+1+1)(k+1+2)(k+1+3)4=(k+1)(k+1+1)(k+1+2)(k+1+3)4
Therefore, P(k+1)P(k+1) holds whenever P(k)P(k) holds.
Hence, the given equality is true for all natural numbers i.e., NN by the principle of mathematical induction.

5. Prove the following by using the principle of mathematical induction for all
n∈Nn∈N :
1.3+2.32+3.33+…+n.3n=(2n-1)3n+1+341.3+2.32+3.33+…+n.3n=(2n-1)3n+1+34

Ans: Let us denote the given equality by P(n),P(n), i.e.,
P(n):1.3+2.32+3.33+…+n3n=(2n-1)3n+1+34P(n):1.3+2.32+3.33+…+n3n=(2n-1)3n+1+34
For n=1n=1,
L.H.S.=1.31=3L.H.S.=1.31=3
R.H.S.=(2.1-1)31+1+34=32+34=124=3R.H.S.=(2.1-1)31+1+34=32+34=124=3
Therefore, P(n)P(n) is true for n=1n=1.
Let us assume that P(k)P(k) is true for some positive integer kk, i.e.,
1.3+2.32+3.33+…+k3k=(2k-1)3k+1+34…(i)1.3+2.32+3.33+…+k3k=(2k-1)3k+1+34…(i)
Now, we have to prove that P(k+1)P(k+1) is also true.
Consider
1.3+2.32+3.33+…+k.3k+(k+1).3k+11.3+2.32+3.33+…+k.3k+(k+1).3k+1
=(1.3+2.32+3.33+…+k.3k)+(k+1).3k+1=(1.3+2.32+3.33+…+k.3k)+(k+1).3k+1
=(2k-1)3k+1+34+(k+1)3k-1=(2k-1)3k+1+34+(k+1)3k-1 Using(i)Using(i)
=(2k-1)3k+1+3+4(k+1)3k+14=(2k-1)3k+1+3+4(k+1)3k+14
=3k+1{2k-1+4(k+1)}+34=3k+1{2k-1+4(k+1)}+34
=3k+1{6k+3}+34=3k+1{6k+3}+34
=3k+1.3{2k+1}+34=3k+1.3{2k+1}+34
=3(k+1)+1{2k+1}+34=3(k+1)+1{2k+1}+34
={2(k+1)-1}3(k+1)+1+34={2(k+1)-1}3(k+1)+1+34
Therefore, P(k+1)P(k+1) holds whenever P(k)P(k) holds.
Hence, the given equality is true for all natural numbers i.e., NN by the principle of mathematical induction.

6. Prove the following by using the principle of mathematical induction for all n∈Nn∈N:
1.2+2.3+3.4+…+n.(n+1)=[n(n+1)(n+2)3]1.2+2.3+3.4+…+n.(n+1)=[n(n+1)(n+2)3]

Ans: Let us denote the given equality by P(n),P(n), i.e.,
P(n):1.2+2.3+3.4+…+n.(n+1)=[n(n+1)(n+2)3]P(n):1.2+2.3+3.4+…+n.(n+1)=[n(n+1)(n+2)3]
For n=1n=1,
L.H.S.=1.2=2L.H.S.=1.2=2
R.H.S.=1(1+1)(1+2)3=1.2.33=2R.H.S.=1(1+1)(1+2)3=1.2.33=2
Therefore, P(n)P(n) is true for n=1n=1.
Let us assume that P(k)P(k) is true for some positive integer kk, i.e.,
1.2+2.3+3.4+…+k.(k+1)=[k(k+1)(k+2)3]…(i)1.2+2.3+3.4+…+k.(k+1)=[k(k+1)(k+2)3]…(i)
Now, we have to prove that P(k+1)P(k+1) is also true.
Consider
1.2+2.3+3.4+…+k.(k+1)+(k+1).(k+2)1.2+2.3+3.4+…+k.(k+1)+(k+1).(k+2)
=[1.2+2.3+3.4+…+k.(k+1)]+(k+1).(k+2)=[1.2+2.3+3.4+…+k.(k+1)]+(k+1).(k+2)
=k(k+1)(k+2)3+(k+1)(k+2)=k(k+1)(k+2)3+(k+1)(k+2) Using(i)Using(i)
=(k+1)(k+2)(k3+1)=(k+1)(k+2)(k3+1)
=(k+1)(k+2)(k+3)3=(k+1)(k+2)(k+3)3
=(k+1)(k+1+1)(k+1+2)3=(k+1)(k+1+1)(k+1+2)3
Therefore, P(k+1)P(k+1) holds whenever P(k)P(k) holds.
Hence, the given equality is true for all natural numbers i.e., NN by the principle of mathematical induction.

7. Prove the following by using the principle of mathematical induction for all n∈Nn∈N:
1.3+3.5+5.7+…+(2n-1)(2n+1)=n(4n2+6n-1)31.3+3.5+5.7+…+(2n-1)(2n+1)=n(4n2+6n-1)3

Ans: Let us denote the given equality by P(n),P(n), i.e.,
P(n):1.3+3.5+5.7+…+(2n-1)(2n+1)=n(4n2+6n-1)3P(n):1.3+3.5+5.7+…+(2n-1)(2n+1)=n(4n2+6n-1)3
For n=1n=1,
L.H.S.=1.3=3L.H.S.=1.3=3
R.H.S.=1(4.12+6.1-1)3=4+6-13=93=3R.H.S.=1(4.12+6.1-1)3=4+6-13=93=3
Therefore, P(n)P(n) is true for n=1n=1.
Let us assume that P(k)P(k) is true for some positive integer kk, i.e.,
1.3+3.5+5.7+…+(2k-1)(2k+1)=k(4k2+6k-1)3…(i)1.3+3.5+5.7+…+(2k-1)(2k+1)=k(4k2+6k-1)3…(i)
Now, we have to prove that P(k+1)P(k+1) is also true.
Consider
(1.3+3.5+5.7+…+(2k-1)(2k+1))+{(k+1)-1}{2(k+1)+1}(1.3+3.5+5.7+…+(2k-1)(2k+1))+{(k+1)-1}{2(k+1)+1}
=k(4k2+6k-1)3+(2k+2-1)(2k+2+1)=k(4k2+6k-1)3+(2k+2-1)(2k+2+1) Using(i)Using(i)
=k(4k2+6k-1)3+(2k+1)(2k+3)=k(4k2+6k-1)3+(2k+1)(2k+3)
=k(4k2+6k-1)3+(4k2+8k+3)=k(4k2+6k-1)3+(4k2+8k+3)
=k(4k2+6k-1)+3(4k2+8k+3)3=k(4k2+6k-1)+3(4k2+8k+3)3
=4k3+6k2-k+12k2+24k+93=4k3+6k2-k+12k2+24k+93
=4k3+18k2+23k+93=4k3+18k2+23k+93
=4k3+14k2+9k+4k2+14k+93=4k3+14k2+9k+4k2+14k+93
=k(4k2+14k+9)+1(4k2+14k+9)3=k(4k2+14k+9)+1(4k2+14k+9)3
=(k+1)(4k2+14k+9)3=(k+1)(4k2+14k+9)3
=(k+1){4k2+8k+4+6k+6-1}3=(k+1){4k2+8k+4+6k+6-1}3
=(k+1){4(k2+2k+1)+6(k+1)-1}3=(k+1){4(k2+2k+1)+6(k+1)-1}3
=(k+1){4(k+1)2+6(k+1)-1}3=(k+1){4(k+1)2+6(k+1)-1}3
Therefore, P(k+1)P(k+1) holds whenever P(k)P(k) holds.
Hence, the given equality is true for all natural numbers i.e., NN by the principle of mathematical induction.

8. Prove the following by using the principle of mathematical induction for all n∈Nn∈N:
1.2+2.22+3.22+….+n.2n=(n-1)2n+1+21.2+2.22+3.22+….+n.2n=(n-1)2n+1+2

Ans: Let us denote the given equality by P(n),P(n), i.e.,
P(n):1.2+2.22+3.22+….+n.2n=(n-1)2n+1+2P(n):1.2+2.22+3.22+….+n.2n=(n-1)2n+1+2
For n=1n=1,
L.H.S.=1.2=2L.H.S.=1.2=2
R.H.S.=(1-1)21+1+2=0+2=2R.H.S.=(1-1)21+1+2=0+2=2
Therefore, P(n)P(n) is true for n=1n=1.
Let us assume that P(k)P(k) is true for some positive integer kk, i.e.,
1.2+2.22+3.22+…+k.2k=(k-1)2k+1+2…(i)1.2+2.22+3.22+…+k.2k=(k-1)2k+1+2…(i)
Now, we have to prove that P(k+1)P(k+1) is also true.
Consider
{1.2+2.22+3.22+…+k.2k}+(k+1).2k+1{1.2+2.22+3.22+…+k.2k}+(k+1).2k+1
=(k-1)2k+1+2+(k+1)2k+1=(k-1)2k+1+2+(k+1)2k+1
=2k+1{(k-1)+(k+1)}+2=2k+1{(k-1)+(k+1)}+2
=2k+1.2k+2=2k+1.2k+2
=k.2(k+1)+1+2=k.2(k+1)+1+2
={(k+1)-1}2(k+1)+1+2={(k+1)-1}2(k+1)+1+2
Therefore, P(k+1)P(k+1) holds whenever P(k)P(k) holds.
Hence, the given equality is true for all natural numbers i.e., NN by the principle of mathematical induction.

9. Prove the following by using the principle of mathematical induction for all n∈Nn∈N:
12+14+18+…+12n=1-12n12+14+18+…+12n=1-12n

Ans: Let us denote the given equality by P(n),P(n), i.e.,
P(n):12+14+18+…+12n=1-12nP(n):12+14+18+…+12n=1-12n
For n=1n=1,
L.H.S.=12L.H.S.=12
R.H.S.=1-121=12R.H.S.=1-121=12
Therefore, P(n)P(n) is true for n=1n=1.
Let us assume that P(k)P(k) is true for some positive integer kk, i.e.,
12+14+18+…+12k=1-12k…(i)12+14+18+…+12k=1-12k…(i)

Now, we have to prove that P(k+1)P(k+1) is also true.
Consider
(12+14+18+…+12k)+12k+1(12+14+18+…+12k)+12k+1
=(1-12k)+12k+1=(1-12k)+12k+1 Using(i)Using(i)
1-12k(1-12)1-12k(1-12)
=1-12k(12)=1-12k(12)
Therefore, P(k+1)P(k+1) holds whenever P(k)P(k) holds.
Hence, the given equality is true for all natural numbers i.e., NN by the principle of mathematical induction.

10. Prove the following by using the principle of mathematical induction for all n∈Nn∈N:
12.5+15.8+18.11+…+1(3n-1)(3n+2)=n(6n+4)12.5+15.8+18.11+…+1(3n-1)(3n+2)=n(6n+4)

Ans: Let us denote the given equality by P(n),P(n), i.e.,
P(n):12.5+15.8+18.11+…+1(3n-1)(3n+2)=n(6n+4)P(n):12.5+15.8+18.11+…+1(3n-1)(3n+2)=n(6n+4)
For n=1n=1,
L.H.S.=12.5=110L.H.S.=12.5=110
R.H.S.=16.1+4=110R.H.S.=16.1+4=110
Therefore, P(n)P(n) is true for n=1n=1.
Let us assume that P(k)P(k) is true for some positive integer kk, i.e.,
12.5+15.8+18.11+…+1(3k+1)(3k+2)=k(6k+4)…(i)12.5+15.8+18.11+…+1(3k+1)(3k+2)=k(6k+4)…(i)
Now, we have to prove that P(k+1)P(k+1) is also true.
Consider
12.5+15.8+18.11+…+1(3k+1)(3k+2)+1{3(k+1)-1}{3(k+1)+2}12.5+15.8+18.11+…+1(3k+1)(3k+2)+1{3(k+1)-1}{3(k+1)+2}
=k6k+4+1(3k+3-1)(3k+3+2)=k6k+4+1(3k+3-1)(3k+3+2) Using(i)Using(i)
=k6k+4+1(3k+2)(3k+5)=k6k+4+1(3k+2)(3k+5)
=k2(3k+2)+1(3k+2)(3k+5)=k2(3k+2)+1(3k+2)(3k+5)
=1(3k+2)(k2+13k+5)=1(3k+2)(k2+13k+5)

=1(3k+2)(k(3k+5)+22(3k+5))=1(3k+2)(k(3k+5)+22(3k+5))

=1(3k+2)(3k2+5k+22(3k+5))=1(3k+2)(3k2+5k+22(3k+5))
=1(3k+2)((3k+2)(k+1)2(3k+5))=1(3k+2)((3k+2)(k+1)2(3k+5))
=(k+1)6k+10=(k+1)6k+10
=(k+1)6(k+1)+4=(k+1)6(k+1)+4
Therefore, P(k+1)P(k+1) holds whenever P(k)P(k) holds.
Hence, the given equality is true for all natural numbers i.e., NN by the principle
of mathematical induction.

11. Prove the following by using the principle of mathematical induction for all n∈Nn∈N:
11.2.3+12.3.4+13.4.5+…+1n(n+1)(n+2)=n(n+3)4(n+1)(n+2)11.2.3+12.3.4+13.4.5+…+1n(n+1)(n+2)=n(n+3)4(n+1)(n+2)

Ans: Let us denote the given equality by P(n),P(n), i.e.,
P(n):11.2.3+12.3.4+13.4.5+…+1n(n+1)(n+2)=n(n+3)4(n+1)(n+2)P(n):11.2.3+12.3.4+13.4.5+…+1n(n+1)(n+2)=n(n+3)4(n+1)(n+2)
For n=1n=1,
L.H.S.=11.2.3=16L.H.S.=11.2.3=16
R.H.S.=1.(1+3)4(1+1)(1+2)=1.44.2.3=16R.H.S.=1.(1+3)4(1+1)(1+2)=1.44.2.3=16
Therefore, P(n)P(n) is true for n=1n=1.
Let us assume that P(k)P(k) is true for some positive integer kk, i.e.,
11.2.3+12.3.4+13.4.5+…+1k(k+1)(k+2)=k(k+3)4(k+1)(k+2)…(i)11.2.3+12.3.4+13.4.5+…+1k(k+1)(k+2)=k(k+3)4(k+1)(k+2)…(i)
Now, we have to prove that P(k+1)P(k+1) is also true.
Consider
[11.2.3+12.3.4+13.4.5+…+1k(k+1)(k+2)]+1(k+1)(k+2)(k+3)[11.2.3+12.3.4+13.4.5+…+1k(k+1)(k+2)]+1(k+1)(k+2)(k+3)
=k(k+3)4(k+1)(k+2)+1(k+1)(k+2)(k+3)=k(k+3)4(k+1)(k+2)+1(k+1)(k+2)(k+3) Using(i)Using(i)
=1(k+1)(k+2){k(k+3)4+1k+3}=1(k+1)(k+2){k(k+3)4+1k+3}
=1(k+1)(k+2){k(k+3)2+44(k+3)}=1(k+1)(k+2){k(k+3)2+44(k+3)}
=1(k+1)(k+2){k(k2+6k+9)+44(k+3)}=1(k+1)(k+2){k(k2+6k+9)+44(k+3)}
=1(k+1)(k+2){k3+6k2+9k+44(k+3)}=1(k+1)(k+2){k3+6k2+9k+44(k+3)}
=1(k+1)(k+2){k3+2k2+k+4k2+8k+44(k+3)}=1(k+1)(k+2){k3+2k2+k+4k2+8k+44(k+3)}
=1(k+1)(k+2){k(k2+2k+1)+4(k2+2k+1)4(k+3)}=1(k+1)(k+2){k(k2+2k+1)+4(k2+2k+1)4(k+3)}
=1(k+1)(k+2){k(k+1)2+4(k+1)24(k+3)}=1(k+1)(k+2){k(k+1)2+4(k+1)24(k+3)}
=(k+1)2(k+4)4(k+1)(k+2)(k+3)=(k+1)2(k+4)4(k+1)(k+2)(k+3)
=(k+1){(k+1)+3}4{(k+1)+1}{(k+1)+2}=(k+1){(k+1)+3}4{(k+1)+1}{(k+1)+2}
Therefore, P(k+1)P(k+1) holds whenever P(k)P(k) holds.
Hence, the given equality is true for all natural numbers i.e., NN by the principle of mathematical induction.

12. Prove the following by using the principle of mathematical induction for all n∈Nn∈N:
a+ar+ar2+…+arn-1=a(rn-1)r-1a+ar+ar2+…+arn-1=a(rn-1)r-1

Ans: Let us denote the given equality by P(n),P(n), i.e.,
P(n):a+ar+ar2+…+arn-1=a(rn-1)r-1P(n):a+ar+ar2+…+arn-1=a(rn-1)r-1
For n=1n=1,
L.H.S.=aL.H.S.=a
R.H.S.=a(r1-1)(r-1)=aR.H.S.=a(r1-1)(r-1)=a
Therefore, P(n)P(n) is true for n=1n=1.
Let us assume that P(k)P(k) is true for some positive integer kk, i.e.,
a+ar+ar2+…+ark-1=a(rk-1)r-1…(i)a+ar+ar2+…+ark-1=a(rk-1)r-1…(i)
Now, we have to prove that P(k+1)P(k+1) is also true.
Consider
{a+ar+ar2+…+ark-1}+ar(k+1)-1{a+ar+ar2+…+ark-1}+ar(k+1)-1
=a(rk-1)r-1+ark=a(rk-1)r-1+ark Using(i)Using(i)
=a(rk-1)+ark(r-1)r-1=a(rk-1)+ark(r-1)r-1
=a(rk-1)+ark+1-arkr-1=a(rk-1)+ark+1-arkr-1
=ark-a+ark+1-arkr-1=ark-a+ark+1-arkr-1
=ark+1-ar-1=ark+1-ar-1
=a(rk+1-1)r-1=a(rk+1-1)r-1
Therefore, P(k+1)P(k+1) holds whenever P(k)P(k) holds.
Hence, the given equality is true for all natural numbers i.e., NN by the principle of mathematical induction.

13. Prove the following by using the principle of mathematical induction for all n∈Nn∈N:
(1+31)(1+54)(1+79)…(1+(2n+1)n2)=(n+1)2(1+31)(1+54)(1+79)…(1+(2n+1)n2)=(n+1)2

Ans: Let us denote the given equality by P(n),P(n), i.e.,
P(n):(1+31)(1+54)(1+79)…(1+(2n+1)n2)=(n+1)2P(n):(1+31)(1+54)(1+79)…(1+(2n+1)n2)=(n+1)2

For n=1n=1,
L.H.S.=(1+31)=4L.H.S.=(1+31)=4
R.H.S.=(1+1)2=22=4R.H.S.=(1+1)2=22=4
Therefore, P(n)P(n) is true for n=1n=1.
Let us assume that P(k)P(k) is true for some positive integer kk, i.e.,
(1+31)(1+54)(1+79)…(1+(2k+1)k2)=(k+1)2…(i)(1+31)(1+54)(1+79)…(1+(2k+1)k2)=(k+1)2…(i)

Now, we have to prove that P(k+1)P(k+1) is also true.
Consider
[(1+31)(1+54)(1+79)…(1+(2k+1)k2)]{1+{2(k+1)+1}(k+1)2}[(1+31)(1+54)(1+79)…(1+(2k+1)k2)]{1+{2(k+1)+1}(k+1)2}
=(k+1)2(1+2(k+1)+1(k+1)2)=(k+1)2(1+2(k+1)+1(k+1)2) Using(i)Using(i)
=(k+1)2[(k+1)2+2(k+1)+1(k+1)2]=(k+1)2[(k+1)2+2(k+1)+1(k+1)2]
=(k+1)2+2(k+1)+1=(k+1)2+2(k+1)+1
={(k+1)+1}2={(k+1)+1}2
Therefore, P(k+1)P(k+1) holds whenever P(k)P(k) holds.
Hence, the given equality is true for all natural numbers i.e., NN by the principle of mathematical induction.

14. Prove the following by using the principle of mathematical induction for all n∈Nn∈N:
(1+11)(1+12)(1+13)…(1+1n)=(n+1)(1+11)(1+12)(1+13)…(1+1n)=(n+1)

Ans: Let us denote the given equality by P(n),P(n), i.e.,
P(n):(1+11)(1+12)(1+13)…(1+1n)=(n+1)P(n):(1+11)(1+12)(1+13)…(1+1n)=(n+1)
For n=1n=1,
L.H.S.=(1+11)=2L.H.S.=(1+11)=2
R.H.S.=(1+1)=2R.H.S.=(1+1)=2
Therefore, P(n)P(n) is true for n=1n=1.
Let us assume that P(k)P(k) is true for some positive integer kk, i.e.,
P(k):(1+11)(1+12)(1+13)…(1+1n)=(k+1)…(i)P(k):(1+11)(1+12)(1+13)…(1+1n)=(k+1)…(i)
Now, we have to prove that P(k+1)P(k+1) is also true.
Consider
[P(k):(1+11)(1+12)(1+13)…(1+1k)](1+1k+1)[P(k):(1+11)(1+12)(1+13)…(1+1k)](1+1k+1)
=(k+1)(1+1k+1)=(k+1)(1+1k+1) Using(i)Using(i)
=(k+1)[(k+1)+1(k+1)]=(k+1)[(k+1)+1(k+1)]
=(k+1)+1=(k+1)+1
Therefore, P(k+1)P(k+1) holds whenever P(k)P(k) holds.
Hence, the given equality is true for all natural numbers i.e., NN by the principle of mathematical induction.

15. Prove the following by using the principle of mathematical induction for all n∈Nn∈N:
12+32+52+…+(2n-1)2=n(2n-1)(2n+1)312+32+52+…+(2n-1)2=n(2n-1)(2n+1)3

Ans: Let us denote the given equality by P(n),P(n), i.e.,
P(n):12+32+52+…+(2n-1)2=n(2n-1)(2n+1)3P(n):12+32+52+…+(2n-1)2=n(2n-1)(2n+1)3
For n=1n=1,
L.H.S.=12=1L.H.S.=12=1
R.H.S.=1(2.1-1)(2.1+1)3=1.1.33=1R.H.S.=1(2.1-1)(2.1+1)3=1.1.33=1
Therefore, P(n)P(n) is true for n=1n=1.
Let us assume that P(k)P(k) is true for some positive integer kk, i.e.,
P(k):12+32+52+…+(2n-1)2=k(2k-1)(2k+1)3…(i)P(k):12+32+52+…+(2n-1)2=k(2k-1)(2k+1)3…(i)
Now, we have to prove that P(k+1)P(k+1) is also true.
Consider
{12+32+52+…+(2k-1)2}+{2(k+1)-1}2{12+32+52+…+(2k-1)2}+{2(k+1)-1}2
=k(2k-1)(2k+1)3+(2k+2-1)2=k(2k-1)(2k+1)3+(2k+2-1)2 Using(i)Using(i)
=k(2k-1)(2k+1)3+(2k+1)2=k(2k-1)(2k+1)3+(2k+1)2
=2(2k-1)(2k+1)+3(2k+1)23=2(2k-1)(2k+1)+3(2k+1)23
=(2k+1){k(2k-1)+3(2k+1)}3=(2k+1){k(2k-1)+3(2k+1)}3
=(2k+1){2k2-k+6k+3}3=(2k+1){2k2-k+6k+3}3
=(2k+1){2k2+5k+3}3=(2k+1){2k2+5k+3}3
=(2k+1){2k2+2k+3k+3}3=(2k+1){2k2+2k+3k+3}3
=(2k+1){2k(k+1)+3(k+1)}3=(2k+1){2k(k+1)+3(k+1)}3
=(2k+1)(k+1)(2k+3)3=(2k+1)(k+1)(2k+3)3
=(k+1){2(k+1)-1}{2(k+1)+1}3=(k+1){2(k+1)-1}{2(k+1)+1}3
Therefore, P(k+1)P(k+1) holds whenever P(k)P(k) holds.
Hence, the given equality is true for all natural numbers i.e., NN by the principle of mathematical induction.

16. Prove the following by using the principle of mathematical induction for all n∈Nn∈N:
11.4+14.7+17.10+…+1(3n-2)(3n+1)=n(3n+1)11.4+14.7+17.10+…+1(3n-2)(3n+1)=n(3n+1)

Ans: Let us denote the given equality by P(n),P(n), i.e.,
P(n):11.4+14.7+17.10+…+1(3n-2)(3n+1)=n(3n+1)P(n):11.4+14.7+17.10+…+1(3n-2)(3n+1)=n(3n+1)
For n=1n=1,
L.H.S.=11.4=14L.H.S.=11.4=14
R.H.S.=13.1+1=14R.H.S.=13.1+1=14
Therefore, P(n)P(n) is true for n=1n=1.
Let us assume that P(k)P(k) is true for some positive integer kk, i.e.,
P(k):11.4+14.7+17.10+…+1(3k-2)(3k+1)=k(3k+1)…(i)P(k):11.4+14.7+17.10+…+1(3k-2)(3k+1)=k(3k+1)…(i)
Now, we have to prove that P(k+1)P(k+1) is also true.
Consider
{11.4+14.7+17.10+…+1(3k-2)(3k+1)}+1{3(k+1)-2}{3(k+1)+1}{11.4+14.7+17.10+…+1(3k-2)(3k+1)}+1{3(k+1)-2}{3(k+1)+1}
=k3k+1+1(3k+1)(3k+4)=k3k+1+1(3k+1)(3k+4) Using(i)Using(i)
=1(3k+1){k+1(3k+4)}=1(3k+1){k+1(3k+4)}
=1(3k+1){k(3k+4)+1(3k+4)}=1(3k+1){k(3k+4)+1(3k+4)}
=1(3k+1){3k2+4k+1(3k+4)}=1(3k+1){3k2+4k+1(3k+4)}
=1(3k+1){3k2+3k+k+1(3k+4)}=1(3k+1){3k2+3k+k+1(3k+4)}
=(3k+1)(k+1)(3k+1)(3k+4)=(3k+1)(k+1)(3k+1)(3k+4)
=(k+1)3(k+1)+1=(k+1)3(k+1)+1
Therefore, P(k+1)P(k+1) holds whenever P(k)P(k) holds.
Hence, the given equality is true for all natural numbers i.e., NN by the principle of mathematical induction.

17. Prove the following by using the principle of mathematical induction for all n∈Nn∈N:
13.5+15.7+17.9+…+1(2n+1)(2n+3)=n2(2n+3)13.5+15.7+17.9+…+1(2n+1)(2n+3)=n2(2n+3)

Ans: For n=1n=1,
L.H.S.=13.5L.H.S.=13.5
R.H.S.=13(2.1+3)=13.5R.H.S.=13(2.1+3)=13.5
Therefore, P(n)P(n) is true for n=1n=1.
Let us assume that P(k)P(k) is true for some positive integer kk, i.e.,
P(k):13.5+15.7+17.9+…+1(2k+1)(2k+3)=k2(2k+3)…(i)P(k):13.5+15.7+17.9+…+1(2k+1)(2k+3)=k2(2k+3)…(i)
Now, we have to prove that P(k+1)P(k+1) is also true.
Consider
[13.5+15.7+17.9+…+1(2k+1)(2k+3)]+1{2(k+1)+1}{2(k+1)+3}[13.5+15.7+17.9+…+1(2k+1)(2k+3)]+1{2(k+1)+1}{2(k+1)+3}
=k3(2k+3)+1(2k+3)(2k+5)=k3(2k+3)+1(2k+3)(2k+5) Using(i)Using(i)
=1(2k+3)[k3+1(2k+5)]=1(2k+3)[k3+1(2k+5)]
=1(2k+3)[k(2k+5)+33(2k+5)]=1(2k+3)[k(2k+5)+33(2k+5)]
=1(2k+3)[2k2+5k+33(2k+5)]=1(2k+3)[2k2+5k+33(2k+5)]
=1(2k+3)[2k2+2k+3k+33(2k+5)]=1(2k+3)[2k2+2k+3k+33(2k+5)]
=1(2k+3)[2k(k+1)+3(k+1)3(2k+5)]=1(2k+3)[2k(k+1)+3(k+1)3(2k+5)]
=(k+1)(2k+3)3(2k+3)(2k+5)=(k+1)(2k+3)3(2k+3)(2k+5)
=(k+1)3{2(k+1)+3}=(k+1)3{2(k+1)+3}
Therefore, P(k+1)P(k+1) holds whenever P(k)P(k) holds.
Hence, the given equality is true for all natural numbers i.e., NN by the principle of mathematical induction.

18. Prove the following by using the principle of mathematical induction for all n∈Nn∈N:
1+2+3+…+n18(2n+1)21+2+3+…+n18(2n+1)2

Ans: Let us denote the given equality by P(n),P(n), i.e.,
P(n):1+2+3+…+n18(2n+1)2P(n):1+2+3+…+n18(2n+1)2
For n=1n=1,
118(2.1+1)2=98118(2.1+1)2=98
Therefore, P(n)P(n) is true for n=1n=1.
Let us assume that P(k)P(k) is true for some positive integer kk, i.e.,
1+2+3+…+k18(2k+1)2…(i)1+2+3+…+k18(2k+1)2…(i)
Now, we have to prove that P(k+1)P(k+1) is also true.
Consider
1+2+3+…+k+(k+1)18(2k+1)2+(k+1)1+2+3+…+k+(k+1)18(2k+1)2+(k+1) Using(i)Using(i)
18{(2k+1)2+8(k+1)}18{(2k+1)2+8(k+1)}
18{4k2+4k+1+8k+8}18{4k2+4k+1+8k+8}
18{4k2+12k+9}18{4k2+12k+9}
18(2k+3)218(2k+3)2
18{2(k+1)+1}218{2(k+1)+1}2
Hence, 1+2+3+…+k+(k+1)18(2k+1)2+(k+1)1+2+3+…+k+(k+1)18(2k+1)2+(k+1)
Therefore, P(k+1)P(k+1) holds whenever P(k)P(k) holds.
Hence, the given equality is true for all natural numbers i.e., NN by the principle of mathematical induction.

19. Prove that following by using the principle of mathematical induction for all n∈Nn∈N:
n(n+1)(n+5)n(n+1)(n+5)is a multiple of 33.

Ans: Let us denote the given statement by P(n)P(n), i.e.,
P(n):n(n+1)(n+5)P(n):n(n+1)(n+5), which is a multiple of 33.
For n=1n=1,
1(1+1)(1+5)=121(1+1)(1+5)=12, which is a multiple of 33.
Therefore, P(n)P(n) is true for n=1n=1.
Let us assume that P(k)P(k) is true for some natural number kk, i.e.,
k(k+1)(k+5)k(k+1)(k+5) is a multiple of 33.
∴k(k+1)(k+5)=3m∴k(k+1)(k+5)=3m, where m∈Nm∈N …(i)
Now, we have to prove that P(k+1)P(k+1) is also true whenever P(k)P(k) is true.
Consider
(k+1){(k+1)+1}{(k+1)+5}(k+1){(k+1)+1}{(k+1)+5}
=(k+1)(k+2){(k+1)+5}=(k+1)(k+2){(k+1)+5}
=(k+1)(k+2)(k+5)+(k+1)(k+2)=(k+1)(k+2)(k+5)+(k+1)(k+2)
={k(k+1)(k+5)+2(k+1)(k+5)}+(k+1)(k+2)={k(k+1)(k+5)+2(k+1)(k+5)}+(k+1)(k+2)
=3m+(k+1){2(k+5)+(k+2)}=3m+(k+1){2(k+5)+(k+2)}
=3m+(k+1){2k+10+k+2}=3m+(k+1){2k+10+k+2}
=3m+(k+1){3k+12}=3m+(k+1){3k+12}
=3m+3(k+1){k+4}=3m+3(k+1){k+4}
=3{m+(k+1)(k+4)}=3×q=3{m+(k+1)(k+4)}=3×q, where q={m+(k+1)(k+4)}q={m+(k+1)(k+4)} is some natural number.
Hence, (k+1){(k+1)+1}{(k+1)+5}(k+1){(k+1)+1}{(k+1)+5} is a multiple of 33.
Therefore, P(k+1)P(k+1) holds whenever P(k)P(k) holds.
Hence, the given equality is true for all natural numbers i.e., NN by the principle of mathematical induction.

20. Prove that following by using the principle of mathematical induction for all n∈Nn∈N:
102n-1+1102n-1+1 is divisible by 1111

Ans: Let us denote the given statement by P(n)P(n), i.e.,
P(n):102n-1+1P(n):102n-1+1 is divisible by 1111.
For n=1n=1,
P(1)=102n-1+1=11P(1)=102n-1+1=11 and P(1)P(1) is divisible by 1111.
Therefore, P(n)P(n) is true for n=1n=1.
Let us assume that P(k)P(k) is true for some natural number kk, i.e.,
i.e., 102k-1+1102k-1+1 is divisible by 1111.
∴102k-1+1=11m∴102k-1+1=11m, where m∈Nm∈N…(i)
Now, we have to prove that P(k+1)P(k+1) is also true whenever P(k)P(k) is true.
Consider
102(k+1)-1+1102(k+1)-1+1
=102k+2-1+1=102k+2-1+1
=102k+1+1=102k+1+1
=102(102k-1+1-1)+1=102(102k-1+1-1)+1
=102(102k-1+1)-102+1=102(102k-1+1)-102+1
=102.11m-100+1=102.11m-100+1 Using(i)Using(i)
=100×11m-99=100×11m-99
=11(100m-9)=11(100m-9)
=11r=11r, where r=(100m-9)r=(100m-9) is some natural number
Therefore, 102(k+1)-1+1102(k+1)-1+1 is divisible by 1111.
Therefore, P(k+1)P(k+1) holds whenever P(k)P(k) holds.
Hence, the given equality is true for all natural numbers i.e., NN by the principle of mathematical induction.

21. Prove that following by using the principle of mathematical induction for all n∈Nn∈N:
x2n-y2nx2n-y2n is divisible by x+yx+y.

Ans: Let us denote the given statement by P(n)P(n), i.e.,
P(n):x2n-y2nP(n):x2n-y2n is divisible by x+yx+y.
For n=1n=1,
P(1)=x2×1-y2×1=x2-y2=(x+y)(x-y)P(1)=x2×1-y2×1=x2-y2=(x+y)(x-y), which is clearly divisible by x+yx+y.
Therefore, P(n)P(n) is true for n=1n=1.
Let us assume that P(k)P(k) is true for some natural number kk, i.e.,
x2k-y2kx2k-y2k is divisible by x+yx+y.
∴∴ Let x2k-y2k=m(x+y)x2k-y2k=m(x+y), where m∈Nm∈N…(i)
Now, we have to prove that P(k+1)P(k+1) is also true whenever P(k)P(k) is true.
Consider
x2(k+1)-y2(k+1)x2(k+1)-y2(k+1)

=x2k.x2-y2k.y2=x2k.x2-y2k.y2
=x2(x2k-y2k+y2k)-y2k.y2=x2(x2k-y2k+y2k)-y2k.y2
=x2{m(x+y)+y2k}-y2k.y2=x2{m(x+y)+y2k}-y2k.y2

=m(x+y)x2+y2k.x2-y2k.y2=m(x+y)x2+y2k.x2-y2k.y2

=m(x+y)x2+y2k(x2-y2)=m(x+y)x2+y2k(x2-y2)

=m(x+y)x2+y2k(x+y)(x-y)=m(x+y)x2+y2k(x+y)(x-y)

=(x+y){mx2+y2k(x-y)}=(x+y){mx2+y2k(x-y)}, which is a factor of (x+y)(x+y).
Therefore, P(k+1)P(k+1) holds whenever P(k)P(k) holds.
Hence, the given equality is true for all natural numbers i.e., NN by the principle of mathematical induction.

22. Prove that following by using the principle of mathematical induction for all n∈Nn∈N:
32n+2-8n-932n+2-8n-9 is divisible by 88.

Ans: Let us denote the given statement by P(n)P(n), i.e.,
P(n):32n+2-8n-9P(n):32n+2-8n-9 is divisible by 88.
For n=1n=1,
P(n)=32×1+2-8×1-9=64P(n)=32×1+2-8×1-9=64, which is divisible by 88.
Therefore, P(n)P(n) is true for n=1n=1.
Let us assume that P(k)P(k) is true for some natural number kk, i.e.,
32k+2-8k-932k+2-8k-9 is divisible by 88.
∴32k+2-8k-9=8m∴32k+2-8k-9=8m; where m∈Nm∈N…(i)
Now, we have to prove that P(k+1)P(k+1) is also true whenever P(k)P(k) is true.
Consider
32(k+1)+2-8(k+1)-932(k+1)+2-8(k+1)-9
32k+2.32-8k-8-932k+2.32-8k-8-9
=32(32k+2-8k-9+8k+9)-8k-17=32(32k+2-8k-9+8k+9)-8k-17
=32(32k+2-8k-9)+32(8k+9)-8k-17=32(32k+2-8k-9)+32(8k+9)-8k-17
=9.8m+9(8k+9)-8k-17=9.8m+9(8k+9)-8k-17
=9.8m+72k+81-8k-17=9.8m+72k+81-8k-17
=9.8m+64k+64=9.8m+64k+64
=8(9m+8k+8)=8(9m+8k+8)
=8r=8r, where r=(9m+8k+8)r=(9m+8k+8) is a natural number
Therefore, 32(k+1)+2-8(k+1)-932(k+1)+2-8(k+1)-9 is divisible by 88.
Therefore, P(k+1)P(k+1) holds whenever P(k)P(k) holds.
Hence, the given equality is true for all natural numbers i.e., NN by the principle of mathematical induction.

23. Prove that following by using the principle of mathematical induction for all n∈Nn∈N:
41n-14n41n-14n is a multiple of 2727.

Ans: Let us denote the given statement by P(n)P(n), i.e.,
P(n):41n-14nP(n):41n-14n is a multiple of 2727.
It can be observed that P(n)P(n) is true for n=1n=1
For n=1n=1,
P(1)=411−141=27P(1)=411−141=27, which is a multiple of 2727.
Therefore, P(n)P(n) is true for n=1n=1.
Let us assume that P(k)P(k) is true for some natural number kk, i.e.,
41k-14k41k-14k is a multiple of 2727.
∴41k-14k=27m,m∈N∴41k-14k=27m,m∈N …(i)
Now, we have to prove that P(k+1)P(k+1) is also true whenever P(k)P(k) is true.
Consider
41k+1-14k+141k+1-14k+1
=41k.41-14k.14=41k.41-14k.14
=41(41k-14k+14k)-14k.14=41(41k-14k+14k)-14k.14
=41.27m+14k(41-14)=41.27m+14k(41-14)
=41.27m+27.14k=41.27m+27.14k
=27(41m-14k)=27(41m-14k)
=27×r=27×r, where r=(41m-14k)r=(41m-14k) is a natural number.
Therefore, 41k+1-14k+141k+1-14k+1 is a multiple of 2727.
Therefore, P(k+1)P(k+1) holds whenever P(k)P(k) holds.
Hence, the given equality is true for all natural numbers i.e., NN by the principle of mathematical induction.

24. Prove that following by using the principle of mathematical induction for all n∈Nn∈N:
(2n+7)(n+3)2(2n+7)(n+3)2

Ans: Let us denote the given statement by P(n)P(n), i.e.,
P(n):(2n+7)(n+3)2P(n):(2n+7)(n+3)2
For n=1n=1,
P(1)=2.1+7=9(1+3)2=16P(1)=2.1+7=9(1+3)2=16, which is true because 916916.
Therefore, P(n)P(n) is true for n=1n=1.
Let us assume that P(k)P(k) is true for some natural number kk, i.e.,
(2k+7)(k+3)2(2k+7)(k+3)2 …(i)
Now, we have to prove that P(k+1)P(k+1) is also true whenever P(k)P(k) is true.
Consider
{2(k+1)+7}=(2k+7)+2{2(k+1)+7}=(2k+7)+2
∴{2(k+1)+7}=(2k+7)+2(k+3)2+2∴{2(k+1)+7}=(2k+7)+2(k+3)2+2 Using(i)Using(i)
2(k+1)+7k2+6k+9+22(k+1)+7k2+6k+9+2
2(k+1)+7k2+6k+112(k+1)+7k2+6k+11
Now, k2+6k+11k2+8k+16k2+6k+11k2+8k+16
∴2(k+1)+7(k+4)2∴2(k+1)+7(k+4)2
2(k+1)+7{(k+1)+3}22(k+1)+7{(k+1)+3}2
Therefore, P(k+1)P(k+1) holds whenever P(k)P(k) holds.
Hence, the given equality is true for all natural numbers i.e., NN by the principle of mathematical induction.